| The value of\[\sqrt{\frac{(\sqrt{12}-\sqrt{8})(\sqrt{3}+\sqrt{2})}{5+\sqrt{24}}}\]is |
A) \[\sqrt{6}-\sqrt{2}\]
B) \[\sqrt{6}+\sqrt{2}\]
C) \[\sqrt{6}-2\]
D) \[2-\sqrt{6}\]
Correct Answer: C
Solution :
| \[\sqrt{\frac{(12--\sqrt{8})(\sqrt{3}+\sqrt{2})}{5+\sqrt{24}}}=\sqrt{\frac{\sqrt{36}-\sqrt{24}+\sqrt{24}-\sqrt{16}}{5+\sqrt{24}}}\] |
| \[=\sqrt{\frac{6-4}{5+\sqrt{24}}}=\sqrt{\frac{2}{5+2\sqrt{6}}}\] |
| \[=\sqrt{\frac{(2)(5-2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})}}=\sqrt{\frac{2\,\,(5-2\sqrt{6})}{25-24}}\] |
| \[=\sqrt{2(5-2\sqrt{6})}\] |
| \[=\sqrt{2\,\,[{{(\sqrt{3})}^{2}}+{{(\sqrt{2})}^{2}}-2\sqrt{3}\sqrt{2})]}\] |
| \[=\sqrt{2\,{{(\sqrt{3}-\sqrt{2})}^{2}}}\] |
| \[=\sqrt{2}\,(\sqrt{3}-\sqrt{2})=\sqrt{6}-2\] |
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